# A rod of length `L` and mass `M` is attached to a pivot and suspended at an angle `alpha ` from the vertical using a support wire, as shown in the diagram. A lump of clay of mass `m` is fired at...

A rod of length `L` and mass `M` is attached to a pivot and suspended at an angle `alpha ` from the vertical using a support wire, as shown in the diagram. A lump of clay of mass `m` is fired at the end of the rod with a velocity `v` . Just before the clay makes contact with the rod, the wire is cut. Assuming the clay and rod stick together after collision, what is the angular velocity of the system? (treat the lump of clay as a point mass)

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We are going to use conservation of angular momentum for this problem. We are going to need the moment of inertia for a rod rotating about its end. I will derive it here using the formula for the inertia about an objects center of mass.

`I_(cm)=int r^2 dm`

Where `r` is the distance from the axis of rotation.

The moment of inertia through the center of mass for a rod on the x-axis with density `lambda=M/L` is:

`I_(cm)=int x^2 dM=int_(-L/2)^(L/2) x^2 (lambda*dx)=lambda/3 x^3|_(-L/2)^(L/2)=1/12 lambda*L^3=12 ML^2`

The parallel axis theorem states the the moment of inertia about the end of the rod will then be:

`I_(rod)=I_(cm)+Mr^2`

Where `r` is the distance from the center of mass.

`I_(rod)=1/12 ML^2+M(L/2)^2=1/3ML^2`

Now for conservation of momentum. Let the quantities with the subscript c be for the clay.

`L_i=L_f`

`r xx p_c=I_(t o t) omega`

`omega=(r xx p_(c))/I_(t o t)`

`omega=(L*mv*sin(90^@ -alpha))/(I_(rod)+I_(c))`

`omega=(L*mv*cos(alpha))/(1/3ML^2+mL^2)`

`omega=(3mv*cos(alpha))/((M+3m)L)`

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